Howie wants to get a $250 loan with a simple interest rate of 4%.

Consider the equation 2(ax+3)=4x−3(4x+8). For what value of a is there no solution to the equation?

erastovalidia [21]

Answer:

the value of $a$ such that there is no solution to the equation is $a = -4$ .

Step-by-step explanation:

Let first simplify the expression presented on statement. The equation has no solution if and only if $a$ is eliminated in the process and an absurd is the result (i.e. $0 = 7$ ).

$2\cdot (a\cdot x + 3) = 4\cdot x -3\cdot (4\cdot x + 8)$

$2\cdot a\cdot x +6 = 4\cdot x -12\cdot x -24$

$2\cdot a \cdot x+6=-8\cdot x-24$

$2\cdot a\cdot x +8\cdot x = -30$

$2\cdot (a+4)\cdot x = -30$

$(a+4)\cdot x = -15$

To obtain an absurd, we need that $a+4 = 0$ . Hence, the value of $a$ such that there is no solution to the equation is:

$a = -4$

Let prove the certainty of the result. We find that an absurd exist: ( $a = -4$ )

$0 = -15$

5 0

2 years ago

Nathan is itemizing deductions on his federal income tax return and had $5800 in his medical expenses last year. If his AGI was

victus00 [196]

Multiply his AGI by 7.5% to find the amount medical needs to meet before deducting:

46,000 x 0.075 = $3,450

He can deduct any amount over $3450.

Subtract that from his medical total to find the amount he can deduct:

5800 - 3450 = 2350

He can deduct $2,350.

6 0

3 years ago

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Which question is statistical? How old is my grandmother? What is the average age of my friends? What are the ages of my friends

lisabon 2012 [21]

<span>What is the average age of my friends?</span>

4 0

2 years ago

Read 2 more answers

PLEASE HELP. it is in the attachment

juin [17]

Answer:

<JKL=123

Step-by-step explanation:

57+57=114

360-114=246

246/2=123

<JKL=123

3 0

3 years ago

Read 2 more answers

13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult

Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that $\frac{ab}{m}$ is and integer.

As m= vb, then $\frac{m}{b}$ is an integer, lets say $\frac{m}{b}$ =v; and as m=ua, then $\frac{m}{a}$ =u.

So $\frac{ab}{m}$ v= $\frac{ab}{m}$ $\frac{m}{b}$ =a, so $\frac{ab}{m}$ divides a; on the other hand, $\frac{ab}{m}$ u= $\frac{ab}{m}$ $\frac{m}{a}$ =b, so $\frac{ab}{m}$ divides b. From this we can conclude that $\frac{ab}{m}$ is a common divisor of a and b.

4 0

3 years ago